Question

Solve this question please class 12th mathematics :)

Ques: The corner points of the feasible region is determined by the following system of linear inequalities: 2x+y ≤ 10, x+3y ≤ 15 , x,y ≥0 are (0,0),(5,0),(3,4) and (0,5). Let Z=px+qy, where p,q>0. Under what conditions p and q the maximum of Z occurs at both (3,4) and (0,5).​

Answer 1

Answer:

q = 3p

Step-by-step explanation:

Given constraints are,

2x + y ≤ 10

x + 3y ≤  15

x

y ≥ 0

The function is to maximize Z = px + qy

The corner points are (0, 0), (5, 0), (3, 4) and (0, 5).

Given that maximum value of Z occurs at (3,4) and (0,5).

∴ Value of Z at (3,4) = Value of Z at (0,5)

=> p(3) + q(4) = p(0) + q(5)

=> 3p + 4q = 5q

=>  q = 3p

∴ Value of Z will be maximum if q = 3p

Hope it helps!

Answer 2

Step-by-step explanation:

Maximum value of Z occurs at (3,4) and (0,5). Therefore Z (3,4) = Z (5,0)

3p + 4q = 5q → 3p = q